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Work done on an object traveling down a slope

  1. Dec 26, 2012 #1
    1. The problem statement, all variables and given/known data

    I'm reading Feynman's lectures on physics to brush up on material I haven't looked at in a while, and I got confused by one of his statements. He claims that the change in potential energy of an object traveling down a slope can be calculated solely by the change in gravitational potential between the starting and ending point, irrespective of the force of constraint. While this seems intuitively correct to me, I'm confused by his explanation, in which he states that the work by gravity is nonzero (mgΔh) while the work done by the force of constraint is zero. To illustrate this, he provides the figure below, to which I have added an x axis for reference:

    work_zps315ca862.png

    I'm confused because work is f*dr. Clearly, the object travels along the y axis (I forgot to label, but assume perpendicular to x) and the force of gravity has an (entirely) y component, so f*dy is a nonzero number. However, the resultant force is shown to have an x component, and the direction of motion also has an x component, so f*dx is also nonzero. Why is the work done by the resultant force considered to be zero?

    2. Relevant equations

    W = f*dr
    U(grav) = -G(m1m2/r^2) =~ mgh

    3. The attempt at a solution

    Described above.

    Not a homework question per se, but I figured that it resembles a homework question moreso than a "real" physics discussion in the general physics forum.

    Thanks for any help!
     
  2. jcsd
  3. Dec 26, 2012 #2

    cepheid

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    Feynman's statement is that the force of constraint does no work. The force of constraint is always perpendicular to the slope, whereas the direction of motion is always parallel to the slope. Hence, the force is always perpendicular to the displacement, and the work (their dot product) is zero.

    Makes sense?
     
  4. Dec 26, 2012 #3
    Yes and no. In the diagram above, the force of constraint has an x component, as does the direction of motion. Therefore, how can the dot product be zero?

    edit: it can be zero if Fx * dx = -Fy * dy. But why does that equality hold?
     
  5. Dec 26, 2012 #4

    cepheid

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    That relation between the components of the vectors F and dr MUST hold, precisely because these two vectors are always perpendicular to each other, and hence their dot product is always zero. The rest comes from the definition of the dot product:

    $$\mathbf{F} \cdot d\mathbf{r} \equiv F_xdx + F_ydy = 0 $$

    Edit: I use boldface to denote vectors, which is pretty standard.
     
  6. Dec 26, 2012 #5
    A-ha! I get it. Thanks for showing me how to think about it!
     
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